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Questions Related to evs

$6$ ml of a standard soap solution $(1$ ml $= 0.001$ g$)$ of  $CaCO _{3}$ were required in titrating $50$ ml of water to produce a good lather. Its degree of hardness is:

evs experiments with water water and its types hard and soft water what floats - what sinks

  1. $120$ ppm

  2. $100$ ppm

  3. $50$ ppm

  4. $60$ ppm

Show answer
Correct Option: A
Explanation:

Degree of hardness $=\dfrac {6\times 0.001}{0.05} =120$ ppm.

On treatment with cold water, an element (A) reacts readily liberating a colorless, odorless gas (B) and a solution (C). Lithium is reacted with (B) yielding a solid product (D) which effervescence with water to give a strongly basic solution (E). When $CO _2$ gas is bubbled through the solution (C), a white ppt (F) is formed but this redissolved forming solution (G) when more $CO _2$ is passed. The precipitate (F) effervesced when moistened with conc.$HCl$ and give deep red coloration to the Bunsen burner flame. (F) on heating with an excess of carbon at $2000^oC$ give (H).

Solution (G) contains a salt which:

(i) causes permanent hardness of water
(ii) can not be obtained in solid-state
(iii) causes temporary hardness of water
(iv) can be obtained in solid-state


Identify the salt with correct options.

evs experiments with water water and its types hard and soft water what floats - what sinks

  1. (i) and (ii)

  2. (i) and (iv)

  3. (ii) and (iii)

  4. (ii) and (iv)

Show answer
Correct Option: C
Explanation:

The metal A is calcium. It reacts with cold water to form calcium hydroxide (C) and hydrogen gas (B) which is colourless and odourless.
$Ca+2H _2O \rightarrow Ca(OH) _2 + H _2$

Lithium reacts with hydrogen gas (B) to form a solid product LiH (D).
$2Li+H _2 \rightarrow 2LiH$

olid product LiH (D) effervescence with water to give a strongly basic solution (E). 
$LiH + H _2O \rightarrow LiOH+ H _2 $

When $CO _2$ gas is bubbled through calcium hydroxide solution (C) , a white ppt of calcium carbonate (F) is formed but this redissolved forming calcium bicarbonate (G) when more $CO _2$ is passed. 

$Ca(OH) _2 + CO _2 \rightarrow CaCO _3 \rightarrow+H _2O$

$CaCO _3+H _2O+CO _2 \rightarrow Ca(HCO _3) _2$

The precipitate (F) (calcium carbonate) effervesced when moistened with concentrated HCl and give deep red colouration to the Bunsen burner flame
$CaCO _3+2HCl \rightarrow CaCl _2 + H _2O +CO _2$

The deep red colour is the characteristic colour of calcium.
Calcium carbonate (F) on heating with excess of carbon at $2000^oC$ give calcium carbide (H).
$CaCO _3 + 3C \xrightarrow {2000^0C} CaC _2+CO+CO _2$

It is $CaHCO _3$ which cannot be obtained in solid-state and causes temporary hardness. Temporary hardness is due to the presence of calcium hydrogen carbonate.   

Hence Option-C is correct

Calculate the temporary and permanent hardness of water sample having the following the following constituents per litre:
$ Ca(HCO _{3}) _{2} = 162\, mg, MgCl _{2} = 95 =\, mg, $
$ NaCl = 585\, mg, Mg(HCO _{3}) _{2} = 73\, mg, $
$ CaSO _{4} = 136\, mg $ 

evs experiments with water water and its types hard and soft water what floats - what sinks

  1. 200 ppm, 150 ppm

  2. 100 ppm, 150 ppm

  3. 150 ppm, 200 ppm

  4. 150 ppm, 150 ppm

Show answer
Correct Option: C
Explanation:

mole of $ Ca(HCO _{3}) _{2} = \dfrac{162\times 10^{-3}mg}{162\,g/mole} = 1\times 10^{-3}\,moles $


Mole of $ Ca(SO _{4}) = \dfrac{136\times 10^{-3}g}{136\,g/mole} = 1\times 10^{3}\,mole $

Total mole of $ Ca = 2\times 10^{-3}\,mole $

mass of $ CaCO _{3} = 2\times 10^{-3}\times 100 = 0.2\,g $

$ \therefore $ ppm (permanent hardness) $ = \dfrac{6.2}{1000}\times 10^{6} = 200\,ppm $

Mole of $ MgCl _{2} = \dfrac{95\times 10^{-3}}{95} = 1\times 10^{-3}\,mole $ 

Mole Mg $ (HCg) _{2} = \dfrac{73\times 10^{-3}}{146} = 5\times 10^{-4}\,mole $
mole Mg $ = 1.5\times 10^{-4}\,mole $

Mole g $ CaCO _{3} $ (In terms of mg) $ = 1.5\times 10^{-3} $

mass $ = 1.5\times 10^{-3} = 0.150\,g $

ppm (temporary hardness) $ = \dfrac{0.150}{100}\times 106 = 150\,ppm $

Hence, the correct option is $\text{C}$

One litre of a sample of hard water contains 55.5 mg of $CaCl _2$ and 4.75 mg of $MgCl _2$. The total harness in terms of ppm of $CaCO _3$ is :

evs experiments with water water and its types hard and soft water what floats - what sinks

  1. 9 ppm

  2. 10 ppm

  3. 20 ppm

  4. none of these

Show answer
Correct Option: D
Explanation:

$55.5$ mg of $CaCl _2$ in 1 litre of water corresponds to $55.5$ ppm.


$4.75$ mg of $MgCl _2$ in 1 litre of water corresponds to $4.75$ ppm.

Also $1 CaCl _2 = 1 CaCO _3$

$111$ ppm $= 100$ ppm

Hence, $55.5$ ppm $=\dfrac {100}{111} \times 55.5=50$ ppm.

Also $1 MgCl _2 = 1 CaCO _3$

$95$ ppm $= 100$ ppm

Hence, $4.75$ ppm $=\dfrac {95}{111} \times 4.75=5$ ppm.

Hence, the total hardness will be $5$ ppm.

In a permutit, the calcium and magnesium ions of hard water are exchanged by:

evs experiments with water water and its types hard and soft water what floats - what sinks

  1. $CO^{2-} _{3} $and $HCO _3{^-}$ ions of permutit

  2. $Na^{+}$ ions of permutit

  3. $Al^{3+}$ ions of permutit

  4. $Si^{4+}$ ions of permutiti

Show answer
Correct Option: B
Explanation:
Permutit is an artificial zeolite. It is sodium aluminium orthosilicate $Na _{2}Al _{2}Si _{2}O _{8}.xH _{2}O$.
$\underset {\text { permutit}}{Na _{2}Al _{2}Si _{2}O _{8}.xH _{2}O} + Ca^{2+}(or Mg^{2+}) \rightarrow  \underset {\text { exhausted permutit }}{Ca Al _{2}Si _{2}O _{8}.xH _{2}O} + 2Na^{+}$

The molecular formula of a commercial resin used for exchanging ions in water softening is $C 8H _7SO _3 $ (Mol. wt. 206). Water would be the maximum uptake of $Ca^{2+} $ ions by the resin when expressed ____ in mole per gram of resin.

evs experiments with water water and its types hard and soft water what floats - what sinks

  1. $\dfrac{2} {309} $

  2. $\dfrac{1} {412} $

  3. $\dfrac{1} {103} $

  4. $\dfrac{1} {206} $

Show answer
Correct Option: B
Explanation:

The chemical reaction for softening water can be given as:-


$2C _8H _7(SO _3)Na+Ca^{+2}\longrightarrow [C _8H _7(SO _3)] _2Ca+2Na^{+}$

$\therefore$  $2$ moles of resin$\equiv1$ mole of $Ca^{2+}$

$\therefore$  Mass of resin = $206\times 2=412g$

For $412g$ of resin, $1$ mole of $Ca^{2+}$ is required.

$\therefore$  Maximum uptake of $Ca^{2+}$ ions $= \cfrac {1}{412}$ mole/gram of resin.

The formula for permutit or zeolite which is used as softer in ion-exchange method is:

evs experiments with water water and its types hard and soft water what floats - what sinks

  1. $NaAlSiO _{4}$

  2. $NaAlO _{2}$

  3. both $A$ and $B$

  4. $Na _{2}SO _{4}$

Show answer
Correct Option: A
Explanation:
Permutit is an artificial zeolite. It is sodium aluminium orthosilicate $Na _{2}Al _{2}Si _{2}O _{8}.xH _{2}O$.
$\underset {\text { permutit}}{Na _{2}Al _{2}Si _{2}O _{8}.xH _{2}O} + Ca^{2+}(or Mg^{2+}) \rightarrow  \underset {\text { exhausted permutit }}{Ca Al _{2}Si _{2}O _{8}.xH _{2}O} + 2Na^{+}$

Gymnosperms are referred to as "naked seeded plants", because 

evs flowers and fruits phanerogams gymnosperms pictorial feature of plant kingdom

  1. They lack ovule

  2. They lack ovaries

  3. They have no seed coat

  4. The embryo is unprotected

Show answer
Correct Option: B
Explanation:

Gymnosperms are those seed plants in which the seeds remain exposed over the surface of the megasporophylls because the latter are not folded to form pistils and thus lack ovary. Flowers are absent and thus fruits are not formed.

So, the correct answer is 'They lack ovaries'.

Gymnosperms do not bear fruits because they do not have

evs flowers and fruits phanerogams gymnosperms pictorial feature of plant kingdom

  1. Seeds

  2. Ovary

  3. Ovule

  4. Pollination

Show answer
Correct Option: B
Explanation:

Gymnosperms, example. Cycas, Pinus, Cedrus etc, bear reproductive structures in the form of cones. The cones that bear female reproductive structures are called female cones. Each female cone contains spirally arranged megasporophylls, that have ovules directly attached to them,  that is,ovules not enclosed in the ovary or are naked. After fertilization the ovule forms the seed and the ovary forms the fruit. Since, in gymnosperms ovary is  absent fruits are not formed.

So, the correct answer is 'Ovary'

Gymnosperms do not include _____________.

evs flowers and fruits phanerogams gymnosperms pictorial feature of plant kingdom

  1. herbs

  2. shrubs

  3. trees

  4. both (a) and (b)

Show answer
Correct Option: A
Explanation:

Living gymnosperms are predominantly middle sized trees (Cycas) to tall trees (Pinus) and shrubs (Ephedra). Rarely they are woody climbers (Gnetum montanum). There are no herbs in gymnosperms.

So the correct option is A.